1. Problem: Find the angle of rotation about the origin that maps point A(4, 5) to A'(−4, −5). 2. Explanation: Rotation about the origin by angle $\theta$ transforms point $(x, y)$ to $(x', y')$ by: $$x' = x\cos\theta - y\sin\theta$$ $$y' = x\sin\theta + y\cos\theta$$ 3. Substitute $A(4,5)$ and $A'(-4,-5)$: $$-4 = 4\cos\theta - 5\sin\theta$$ $$-5 = 4\sin\theta + 5\cos\theta$$ 4. Multiply first equation by 5 and second by 4: $$-20 = 20\cos\theta - 25\sin\theta$$ $$-20 = 16\sin\theta + 20\cos\theta$$ 5. Subtract second from first: $$0 = 4\cos\theta - 41\sin\theta$$ 6. Rearranged: $$4\cos\theta = 41\sin\theta \implies \tan\theta = \frac{4}{41}$$ 7. Check if this matches the image point. But since $A' = (-4, -5)$ is exactly $-1$ times $A = (4,5)$, this corresponds to rotation by $180^\circ$ or $-180^\circ$. 8. Therefore, the angle of rotation is $\boxed{\pm 180^\circ}$. --- 9. Problem: Find the image of point $P(x,y)$ under translation vector $\begin{pmatrix}m \\ n\end{pmatrix}$. 10. Explanation: Translation adds the vector to the point coordinates: $$P'(x', y') = (x + m, y + n)$$ 11. So the image is $\boxed{(x + m, y + n)}$. --- 12. Problem: Given $P(x,y)$ maps to $P'(m,n)$ under translation, find the translation vector. 13. Explanation: Translation vector is the difference: $$\begin{pmatrix}m - x \\ n - y\end{pmatrix}$$ 14. So the translation vector is $\boxed{\begin{pmatrix}m - x \\ n - y\end{pmatrix}}$. --- 15. Problem: Find the translation vector if point $(4,5)$ maps to $(6,8)$ under translation. 16. Calculation: $$\begin{pmatrix}6 - 4 \\ 8 - 5\end{pmatrix} = \begin{pmatrix}2 \\ 3\end{pmatrix}$$ 17. So the translation vector is $\boxed{\begin{pmatrix}2 \\ 3\end{pmatrix}}$. --- 18. Problem: Find coordinates of image of $A(x,y)$ enlarged about origin with scale factor $K$. 19. Explanation: Enlargement about origin scales coordinates: $$A'(x', y') = (Kx, Ky)$$ 20. So the image coordinates are $\boxed{(Kx, Ky)}$. --- 21. Problem: Find image of $A(4,5)$ enlarged by $E[(0,1), 2]$. 22. Explanation: Enlargement with center $(0,1)$ and scale factor 2: $$A' = (0 + 2(4 - 0), 1 + 2(5 - 1)) = (8, 9)$$ 23. So the image coordinates are $\boxed{(8, 9)}$. --- 24. Problem: Find matrix $B$ such that $AB = I_2$ for $$A = \begin{pmatrix}2 & 1 \\ 3 & 4\end{pmatrix}$$ 25. Explanation: $B = A^{-1}$, inverse of $A$. 26. Calculate determinant: $$\det A = 2 \times 4 - 1 \times 3 = 8 - 3 = 5$$ 27. Inverse: $$B = \frac{1}{5} \begin{pmatrix}4 & -1 \\ -3 & 2\end{pmatrix} = \begin{pmatrix} \frac{4}{5} & -\frac{1}{5} \\ -\frac{3}{5} & \frac{2}{5} \end{pmatrix}$$ 28. So, $$B = \boxed{\begin{pmatrix} \frac{4}{5} & -\frac{1}{5} \\ -\frac{3}{5} & \frac{2}{5} \end{pmatrix}}$$ --- 29. Problem: Given $$A = \begin{pmatrix}2 & 1 \\ 5 & 3\end{pmatrix}, B = \begin{pmatrix}3 & 4 \\ 2 & 3\end{pmatrix}$$ Find matrix $C$ such that $AC = B$. 30. Explanation: Multiply both sides by $A^{-1}$: $$C = A^{-1} B$$ 31. Calculate $A^{-1}$: $$\det A = 2 \times 3 - 1 \times 5 = 6 - 5 = 1$$ 32. So, $$A^{-1} = \begin{pmatrix}3 & -1 \\ -5 & 2\end{pmatrix}$$ 33. Multiply: $$C = \begin{pmatrix}3 & -1 \\ -5 & 2\end{pmatrix} \begin{pmatrix}3 & 4 \\ 2 & 3\end{pmatrix} = \begin{pmatrix}7 & 9 \\ -11 & -14\end{pmatrix}$$ 34. So, $$C = \boxed{\begin{pmatrix}7 & 9 \\ -11 & -14\end{pmatrix}}$$ --- 35. Problem: Show that for $$A = \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix}, A^2 - 2A = 0$$ 36. Calculate $A^2$: $$A^2 = A \times A = \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix} \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix} = \begin{pmatrix}2 & 2 \\ 2 & 2\end{pmatrix}$$ 37. Calculate $A^2 - 2A$: $$\begin{pmatrix}2 & 2 \\ 2 & 2\end{pmatrix} - 2 \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix} = \begin{pmatrix}2 & 2 \\ 2 & 2\end{pmatrix} - \begin{pmatrix}2 & 2 \\ 2 & 2\end{pmatrix} = \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}$$ 38. Hence, $A^2 - 2A$ is the null matrix. --- 39. Problem: Show that for $$A = \begin{pmatrix}1 & 1 \\ 8 & 3\end{pmatrix}, A^2 - 4A = 5I$$ 40. Calculate $A^2$: $$A^2 = \begin{pmatrix}1 & 1 \\ 8 & 3\end{pmatrix} \begin{pmatrix}1 & 1 \\ 8 & 3\end{pmatrix} = \begin{pmatrix}1+8 & 1+3 \\ 8+24 & 8+9\end{pmatrix} = \begin{pmatrix}9 & 4 \\ 32 & 17\end{pmatrix}$$ 41. Calculate $A^2 - 4A$: $$\begin{pmatrix}9 & 4 \\ 32 & 17\end{pmatrix} - 4 \begin{pmatrix}1 & 1 \\ 8 & 3\end{pmatrix} = \begin{pmatrix}9 & 4 \\ 32 & 17\end{pmatrix} - \begin{pmatrix}4 & 4 \\ 32 & 12\end{pmatrix} = \begin{pmatrix}5 & 0 \\ 0 & 5\end{pmatrix} = 5I$$ 42. Hence, $A^2 - 4A = 5I$ is proven. --- 43. Problem: If $$A = \begin{pmatrix}x & 5 \\ 3 & y\end{pmatrix}, B = \begin{pmatrix}4 & 3 \\ 5 & -2\end{pmatrix}$$ and $A^t = B$, find $x$ and $y$. 44. Explanation: Transpose $A$ is $$A^t = \begin{pmatrix}x & 3 \\ 5 & y\end{pmatrix} = B = \begin{pmatrix}4 & 3 \\ 5 & -2\end{pmatrix}$$ 45. Equate elements: $$x = 4, \quad 3 = 3, \quad 5 = 5, \quad y = -2$$ 46. So, $$\boxed{x=4, y=-2}$$ --- 47. Problem: If $$(a \, b) - (1 \, 2) = (5 \, 6)$$ and $$(c \, d) - (3 \, 4) = (6 \, 8)$$ find $a,b,c,d$. 48. Calculate: $$a - 1 = 5 \implies a = 6$$ $$b - 2 = 6 \implies b = 8$$ $$c - 3 = 6 \implies c = 9$$ $$d - 4 = 8 \implies d = 12$$ 49. So, $$\boxed{a=6, b=8, c=9, d=12}$$ --- 50. Problem: If $$(a \, b) - (2 \, 4) = (10 \, 12)$$ and $$(c \, d) - (6 \, 8) = (12 \, 16)$$ find $a,b,c,d$. 51. Calculate: $$a - 2 = 10 \implies a = 12$$ $$b - 4 = 12 \implies b = 16$$ $$c - 6 = 12 \implies c = 18$$ $$d - 8 = 16 \implies d = 24$$ 52. So, $$\boxed{a=12, b=16, c=18, d=24}$$ --- 53. Problem: Given $$A = \begin{pmatrix}3 \\ 2\end{pmatrix}, B = \begin{pmatrix}-1 \\ 4\end{pmatrix}, C = \begin{pmatrix}-3 \\ -1\end{pmatrix}$$ Find $A + 2B - 3C$. 54. Calculate: $$2B = \begin{pmatrix}-2 \\ 8\end{pmatrix}, -3C = \begin{pmatrix}9 \\ 3\end{pmatrix}$$ 55. Sum: $$A + 2B - 3C = \begin{pmatrix}3 \\ 2\end{pmatrix} + \begin{pmatrix}-2 \\ 8\end{pmatrix} + \begin{pmatrix}9 \\ 3\end{pmatrix} = \begin{pmatrix}10 \\ 13\end{pmatrix}$$ 56. So, $$\boxed{\begin{pmatrix}10 \\ 13\end{pmatrix}}$$ --- 57. Problem: Find $a,b$ if midpoint of $(a+1,1)$ and $(6,b+2)$ is $(4,2)$. 58. Midpoint formula: $$\left(\frac{a+1+6}{2}, \frac{1 + b + 2}{2}\right) = (4,2)$$ 59. Equate: $$\frac{a+7}{2} = 4 \implies a+7=8 \implies a=1$$ $$\frac{b+3}{2} = 2 \implies b+3=4 \implies b=1$$ 60. So, $$\boxed{a=1, b=1}$$ --- 61. Problem: Find other end of line segment if one end is $(4,4)$ and midpoint is $(-2,2)$. 62. Let other end be $(x,y)$. 63. Midpoint formula: $$\left(\frac{4+x}{2}, \frac{4+y}{2}\right) = (-2, 2)$$ 64. Equate: $$\frac{4+x}{2} = -2 \implies 4+x = -4 \implies x = -8$$ $$\frac{4+y}{2} = 2 \implies 4+y = 4 \implies y = 0$$ 65. So, $$\boxed{(-8, 0)}$$ --- 66. Problem: Find other end if one end is $(2,1)$ and midpoint is $(4,2)$. 67. Let other end be $(x,y)$. 68. Midpoint formula: $$\left(\frac{2+x}{2}, \frac{1+y}{2}\right) = (4, 2)$$ 69. Equate: $$\frac{2+x}{2} = 4 \implies 2+x=8 \implies x=6$$ $$\frac{1+y}{2} = 2 \implies 1+y=4 \implies y=3$$ 70. So, $$\boxed{(6, 3)}$$ --- 71. Problem: Find ratio in which point $(5,3)$ divides line segment joining $(2,3)$ and $(7,3)$. 72. Since $y$-coordinates are same, ratio $m:n$ satisfies: $$x = \frac{m \times 7 + n \times 2}{m+n} = 5$$ 73. Let ratio be $m:n = r:1$: $$5 = \frac{7r + 2}{r+1} \implies 5(r+1) = 7r + 2 \implies 5r + 5 = 7r + 2 \implies 2r = 3 \implies r = \frac{3}{2}$$ 74. So ratio is $3:2$. --- 75. Problem: Find ratio in which point $(2,b)$ divides line segment joining $(-4,3)$ and $(6,3)$. 76. Since $y$-coordinates are same, ratio $m:n$ satisfies: $$x = \frac{m \times 6 + n \times (-4)}{m+n} = 2$$ 77. Let ratio be $m:n = r:1$: $$2 = \frac{6r - 4}{r+1} \implies 2(r+1) = 6r - 4 \implies 2r + 2 = 6r - 4 \implies 4r = 6 \implies r = \frac{3}{2}$$ 78. So ratio is $3:2$. --- 79. Problem: Find ratio in which X-axis divides line joining $(2,-4)$ and $(-3,6)$. 80. Let ratio be $m:n = r:1$ and point on X-axis has $y=0$. 81. Using section formula for $y$: $$0 = \frac{m \times 6 + n \times (-4)}{m+n} = \frac{6r - 4}{r+1}$$ 82. Solve: $$6r - 4 = 0 \implies 6r = 4 \implies r = \frac{2}{3}$$ 83. So ratio is $2:3$. --- 84. Problem: Find equation of line with slope $-3$ passing through $(2,-2)$. 85. Use point-slope form: $$y - y_1 = m(x - x_1)$$ $$y + 2 = -3(x - 2)$$ $$y + 2 = -3x + 6$$ $$3x + y = 4$$ 86. So equation is $\boxed{3x + y = 4}$. --- 87. Problem: Find coordinates where line $4x + 5y = 20$ intersects X-axis and Y-axis. 88. For X-axis, $y=0$: $$4x = 20 \implies x=5$$ 89. For Y-axis, $x=0$: $$5y = 20 \implies y=4$$ 90. So, $$A = (5,0), B = (0,4)$$ --- 91. Problem: Given $AP = BP$ and $P=(4,3)$, find coordinates of $A$ and $B$. 92. Since $AP=BP$, $P$ is midpoint of $AB$. 93. Let $A=(x_1,y_1)$ and $B=(x_2,y_2)$. 94. Midpoint $P$: $$\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = (4,3)$$ 95. Given $A$ and $B$ lie on axes, so $A=(a,0)$ and $B=(0,b)$. 96. Substitute: $$\frac{a + 0}{2} = 4 \implies a=8$$ $$\frac{0 + b}{2} = 3 \implies b=6$$ 97. So, $$A = (8,0), B = (0,6)$$ --- 98. Problem: Find $p$ if distance from $(2,-3)$ to line $px - 4y + 7=0$ is 5. 99. Distance formula: $$d = \frac{|p(2) -4(-3) + 7|}{\sqrt{p^2 + (-4)^2}} = 5$$ 100. Simplify numerator: $$|2p + 12 + 7| = |2p + 19|$$ 101. Equation: $$\frac{|2p + 19|}{\sqrt{p^2 + 16}} = 5$$ 102. Square both sides: $$(2p + 19)^2 = 25(p^2 + 16)$$ 103. Expand: $$4p^2 + 76p + 361 = 25p^2 + 400$$ 104. Rearrange: $$0 = 25p^2 - 4p^2 - 76p + 400 - 361$$ $$0 = 21p^2 - 76p + 39$$ 105. Solve quadratic: $$p = \frac{76 \pm \sqrt{(-76)^2 - 4 \times 21 \times 39}}{2 \times 21} = \frac{76 \pm \sqrt{5776 - 3276}}{42} = \frac{76 \pm \sqrt{2500}}{42}$$ 106. So, $$p = \frac{76 \pm 50}{42}$$ 107. Two solutions: $$p = \frac{126}{42} = 3, \quad p = \frac{26}{42} = \frac{13}{21}$$ 108. So, $$\boxed{p = 3 \text{ or } \frac{13}{21}}$$